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When the positive integer x is divided by 11, the quotient is y and the remainder 3. When x is divided by 19, the remainder is also 3. What is the remainder when y is divided by 19?

A.4 B.3 C.2 D.1 E.0

(1) When the positive integer x is divided by 11, the quotient is y and the remainder 3 --> \(x=11y+3\);(2) When x is divided by 19, the remainder is also 3 --> \(x=19q+3\).

Subtract (2) from (1) --> \(19q=11y\) --> \(y=\frac{19q}{11}\). Now as \(y\) and \(q\) are integers and 19 is prime \(\frac{q}{11}\) must be an integer --> \(y=19*integer\) --> \(y\) is a multiple of 19, hence when divide by 19 remainder is 0.

General Discussion

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When the positive integer x is divided by 11, the quotient is y and the remainder 3. When x is divided by 19, the remainder is also 3. What is the remainder when y is divided by 19?

A.4 B.3 C.2 D.1 E.0

IMHO E

We have two cases.. x = 11*y + 3 and x = 19*m + 3... where just like y..its an integer..

equating both the equations..11*y + 3 = 19*m + 3y = 19*m / 11. now both 11 and 19 are prime...so m has to be a multiple of 11.. Only then we can get y as integer.. so let say m= 11*p, where p is a positive integer..

so y = (19 * 11 * p)/11 and when y is divide by 19..we get remainder as zero.

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When the positive integer x is divided by 11, the quotient is y and the remainder 3. When x is divided by 19, the remainder is also 3. What is the remainder when y is divided by 19?

A) 0B) 1C) 2D) 3E) 4

So I got this far:

If x divided by 11 has a quotient of y and a remainder of 3, x can be expressed as x = 11y + 3, where y is an integer (by definition, a quotient is an integer). If x divided by 19 also has a remainder of 3, we can also express x as x = 19z + 3, where z is an integer.

We can set the two equations equal to each other:11y + 3 = 19z + 311y = 19z

This is where I get lost

How does 11y = 19z help me determine what the remainder is when y is divided by 19???? Is there another step somewhere?

Here's the rest of the explanation:

The question asks us what the remainder is when y is divided by 19. From the equation we see that 11y is a multiple of 19 because z is an integer. y itself must be a multiple of 19 since 11, the coefficient of y, is not a multiple of 19.

Answer is A) 0

Please tell me there's an easier way to do this then to "assume" y is a multiple of 19 so then there's no remainder. Thoughts??

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29 Nov 2010, 21:34

both 11 and 19 are prime numbers therefore z and y have to be multiples of these prime numbers in order for 11y=19Z to be true. Since there prime there is no other way possible. And if y is a multiple of Y, then Y/19 will result in a 0 remainder.

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When the positive integer x is divided by 11, the quotient is y and the remainder 3. When x is divided by 19, the remainder is also 3. What is the remainder when y is divided by 19?

A) 0 B) 1 C) 2 D) 3 E) 4

Could you please explain the answer??I am confused by the following explanation:

If x divided by 11 has a quotient of y and a remainder of 3, x can be expressed as x = 11y + 3, where y is an integer (by definition, a quotient is an integer). If x divided by 19 also has a remainder of 3, we can also express x as x = 19z + 3, where z is an integer.

We can set the two equations equal to each other:11y + 3 = 19z + 311y = 19z

The question asks us what the remainder is when y is divided by 19. From the equation we see that 11y is a multiple of 19 because z is an integer. y itself must be a multiple of 19 since 11, the coefficient of y, is not a multiple of 19.

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When the positive integer x is divided by 11, the quotient is y and the remainder 3. When x is divided by 19, the remainder is also 3. What is the remainder when y is divided by 19?

A.4 B.3 C.2 D.1 E.0

(1) When the positive integer x is divided by 11, the quotient is y and the remainder 3 --> \(x=11y+3\);(2) When x is divided by 19, the remainder is also 3 --> \(x=19q+3\).

Subtract (2) from (1) --> \(19q=11y\) --> \(y=\frac{19q}{11}\). Now as \(y\) and \(q\) are integers and 19 is prime \(\frac{q}{11}\) must be an integer --> \(y=19*integer\) --> \(y\) is a multiple of 19, hence when divide by 19 remainder is 0.

Answer: E.

Thanks or the explanation.Hw you figured out q/11 is an integer. I know that q is an integer and 11 also; but cannot guarantee that q/ 11 - will be an integer. Please clarify._________________

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When the positive integer x is divided by 11, the quotient is y and the remainder 3. When x is divided by 19, the remainder is also 3. What is the remainder when y is divided by 19?

A.4 B.3 C.2 D.1 E.0

(1) When the positive integer x is divided by 11, the quotient is y and the remainder 3 --> \(x=11y+3\);(2) When x is divided by 19, the remainder is also 3 --> \(x=19q+3\).

Subtract (2) from (1) --> \(19q=11y\) --> \(y=\frac{19q}{11}\). Now as \(y\) and \(q\) are integers and 19 is prime \(\frac{q}{11}\) must be an integer --> \(y=19*integer\) --> \(y\) is a multiple of 19, hence when divide by 19 remainder is 0.

Answer: E.

Thanks or the explanation.Hw you figured out q/11 is an integer. I know that q is an integer and 11 also; but cannot guarantee that q/ 11 - will be an integer. Please clarify.

Let me ask you a question: how can y be an integer if \(\frac{q}{11}\) is not?
_________________